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Re:Chance of Dying From Backscatter Radiation Same as Chance... by kwantam on Thursday November 18, 2010 @05:28PM (#1919555)
VVVVVV Ahh, ayup, you're correct. Brainfart.
Re:Chance of Dying From Backscatter Radiation Same as Chance... by andrewcooke on Thursday November 18, 2010 @05:51PM (#1919635)
Sorry for being pedantic, but no - these are probabilities not random variates. If the probability of either alone is p then the probability of one or the other or both is:

1 - (1-p)(1-p) = 1 - (1-2p+p^2) = 2p - p^2 ~ 2p

(where the last step is because p^2 << p because p << 1; the basic idea is that it's "certainty minus the probability that neither occur")

Re:Chance of Dying From Backscatter Radiation Same as Chance... by joe_the_user on Thursday November 18, 2010 @06:05PM (#1919675)
Not close to true for two small, independent events.

We can safely assume that the chance of their occurrence together is going to be very small and so the chance of one or the other is approximately the sum of their individual probability.

P(X ∪ Y) = P(X) + P(Y) - P(X ∩ Y) ~ P(X) + P(Y)